3.322 \(\int \frac{x^5 (a+b x^2+c x^4)^{3/2}}{d+e x^2} \, dx\)

Optimal. Leaf size=482 \[ -\frac{\left (16 b c^2 e^3 \left (3 a^2 e^2-3 a b d e+b^2 d^2\right )+6 b^3 c e^4 (b d-4 a e)-384 c^4 d^3 e (b d-a e)+96 c^3 d e^2 (b d-a e)^2+3 b^5 e^5+256 c^5 d^5\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{512 c^{7/2} e^6}+\frac{\left (a+b x^2+c x^4\right )^{3/2} \left (-3 b^2 e^2-6 c e x^2 (b e+2 c d)-6 b c d e+16 c^2 d^2\right )}{96 c^2 e^3}+\frac{\sqrt{a+b x^2+c x^4} \left (-2 c e x^2 \left (-8 c^2 d e (2 b d-3 a e)-6 b c e^2 (b d-2 a e)-3 b^3 e^3+32 c^3 d^3\right )+6 b^2 c e^3 (b d-2 a e)-32 c^3 d^2 e (5 b d-4 a e)+8 b c^2 d e^2 (2 b d-3 a e)+3 b^4 e^4+128 c^4 d^4\right )}{256 c^3 e^5}+\frac{d^2 \left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt{a+b x^2+c x^4} \sqrt{a e^2-b d e+c d^2}}\right )}{2 e^6}+\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 c e} \]

[Out]

((128*c^4*d^4 + 3*b^4*e^4 - 32*c^3*d^2*e*(5*b*d - 4*a*e) + 8*b*c^2*d*e^2*(2*b*d - 3*a*e) + 6*b^2*c*e^3*(b*d -
2*a*e) - 2*c*e*(32*c^3*d^3 - 3*b^3*e^3 - 8*c^2*d*e*(2*b*d - 3*a*e) - 6*b*c*e^2*(b*d - 2*a*e))*x^2)*Sqrt[a + b*
x^2 + c*x^4])/(256*c^3*e^5) + ((16*c^2*d^2 - 6*b*c*d*e - 3*b^2*e^2 - 6*c*e*(2*c*d + b*e)*x^2)*(a + b*x^2 + c*x
^4)^(3/2))/(96*c^2*e^3) + (a + b*x^2 + c*x^4)^(5/2)/(10*c*e) - ((256*c^5*d^5 + 3*b^5*e^5 + 6*b^3*c*e^4*(b*d -
4*a*e) - 384*c^4*d^3*e*(b*d - a*e) + 96*c^3*d*e^2*(b*d - a*e)^2 + 16*b*c^2*e^3*(b^2*d^2 - 3*a*b*d*e + 3*a^2*e^
2))*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(512*c^(7/2)*e^6) + (d^2*(c*d^2 - b*d*e + a*e^
2)^(3/2)*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x^2)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x^2 + c*x^4])])/(
2*e^6)

________________________________________________________________________________________

Rubi [A]  time = 1.10293, antiderivative size = 482, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {1251, 1653, 814, 843, 621, 206, 724} \[ -\frac{\left (16 b c^2 e^3 \left (3 a^2 e^2-3 a b d e+b^2 d^2\right )+6 b^3 c e^4 (b d-4 a e)-384 c^4 d^3 e (b d-a e)+96 c^3 d e^2 (b d-a e)^2+3 b^5 e^5+256 c^5 d^5\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{512 c^{7/2} e^6}+\frac{\left (a+b x^2+c x^4\right )^{3/2} \left (-3 b^2 e^2-6 c e x^2 (b e+2 c d)-6 b c d e+16 c^2 d^2\right )}{96 c^2 e^3}+\frac{\sqrt{a+b x^2+c x^4} \left (-2 c e x^2 \left (-8 c^2 d e (2 b d-3 a e)-6 b c e^2 (b d-2 a e)-3 b^3 e^3+32 c^3 d^3\right )+6 b^2 c e^3 (b d-2 a e)-32 c^3 d^2 e (5 b d-4 a e)+8 b c^2 d e^2 (2 b d-3 a e)+3 b^4 e^4+128 c^4 d^4\right )}{256 c^3 e^5}+\frac{d^2 \left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt{a+b x^2+c x^4} \sqrt{a e^2-b d e+c d^2}}\right )}{2 e^6}+\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 c e} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*x^2 + c*x^4)^(3/2))/(d + e*x^2),x]

[Out]

((128*c^4*d^4 + 3*b^4*e^4 - 32*c^3*d^2*e*(5*b*d - 4*a*e) + 8*b*c^2*d*e^2*(2*b*d - 3*a*e) + 6*b^2*c*e^3*(b*d -
2*a*e) - 2*c*e*(32*c^3*d^3 - 3*b^3*e^3 - 8*c^2*d*e*(2*b*d - 3*a*e) - 6*b*c*e^2*(b*d - 2*a*e))*x^2)*Sqrt[a + b*
x^2 + c*x^4])/(256*c^3*e^5) + ((16*c^2*d^2 - 6*b*c*d*e - 3*b^2*e^2 - 6*c*e*(2*c*d + b*e)*x^2)*(a + b*x^2 + c*x
^4)^(3/2))/(96*c^2*e^3) + (a + b*x^2 + c*x^4)^(5/2)/(10*c*e) - ((256*c^5*d^5 + 3*b^5*e^5 + 6*b^3*c*e^4*(b*d -
4*a*e) - 384*c^4*d^3*e*(b*d - a*e) + 96*c^3*d*e^2*(b*d - a*e)^2 + 16*b*c^2*e^3*(b^2*d^2 - 3*a*b*d*e + 3*a^2*e^
2))*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(512*c^(7/2)*e^6) + (d^2*(c*d^2 - b*d*e + a*e^
2)^(3/2)*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x^2)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x^2 + c*x^4])])/(
2*e^6)

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{x^5 \left (a+b x^2+c x^4\right )^{3/2}}{d+e x^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2 \left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx,x,x^2\right )\\ &=\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 c e}+\frac{\operatorname{Subst}\left (\int \frac{\left (-\frac{5}{2} b d e-\frac{5}{2} e (2 c d+b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx,x,x^2\right )}{10 c e^2}\\ &=\frac{\left (16 c^2 d^2-6 b c d e-3 b^2 e^2-6 c e (2 c d+b e) x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{96 c^2 e^3}+\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 c e}-\frac{\operatorname{Subst}\left (\int \frac{\left (-\frac{5}{4} d e \left (6 b^2 c d e+8 a c^2 d e+3 b^3 e^2-4 b c \left (4 c d^2+3 a e^2\right )\right )+\frac{5}{4} e \left (32 c^3 d^3-3 b^3 e^3-8 c^2 d e (2 b d-3 a e)-6 b c e^2 (b d-2 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{d+e x} \, dx,x,x^2\right )}{80 c^2 e^4}\\ &=\frac{\left (128 c^4 d^4+3 b^4 e^4-32 c^3 d^2 e (5 b d-4 a e)+8 b c^2 d e^2 (2 b d-3 a e)+6 b^2 c e^3 (b d-2 a e)-2 c e \left (32 c^3 d^3-3 b^3 e^3-8 c^2 d e (2 b d-3 a e)-6 b c e^2 (b d-2 a e)\right ) x^2\right ) \sqrt{a+b x^2+c x^4}}{256 c^3 e^5}+\frac{\left (16 c^2 d^2-6 b c d e-3 b^2 e^2-6 c e (2 c d+b e) x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{96 c^2 e^3}+\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 c e}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{5}{8} d e \left (6 b^4 c d e^3+3 b^5 e^4+8 b^3 c e^2 \left (2 c d^2-3 a e^2\right )-16 b^2 c^2 d e \left (10 c d^2+3 a e^2\right )-32 a c^3 d e \left (4 c d^2+5 a e^2\right )+16 b c^2 \left (8 c^2 d^4+20 a c d^2 e^2+3 a^2 e^4\right )\right )-\frac{5}{8} e \left (256 c^5 d^5+3 b^5 e^5+6 b^3 c e^4 (b d-4 a e)-384 c^4 d^3 e (b d-a e)+96 c^3 d e^2 (b d-a e)^2+16 b c^2 e^3 \left (b^2 d^2-3 a b d e+3 a^2 e^2\right )\right ) x}{(d+e x) \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{320 c^3 e^6}\\ &=\frac{\left (128 c^4 d^4+3 b^4 e^4-32 c^3 d^2 e (5 b d-4 a e)+8 b c^2 d e^2 (2 b d-3 a e)+6 b^2 c e^3 (b d-2 a e)-2 c e \left (32 c^3 d^3-3 b^3 e^3-8 c^2 d e (2 b d-3 a e)-6 b c e^2 (b d-2 a e)\right ) x^2\right ) \sqrt{a+b x^2+c x^4}}{256 c^3 e^5}+\frac{\left (16 c^2 d^2-6 b c d e-3 b^2 e^2-6 c e (2 c d+b e) x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{96 c^2 e^3}+\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 c e}+\frac{\left (d^2 \left (c d^2-b d e+a e^2\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{2 e^6}-\frac{\left (256 c^5 d^5+3 b^5 e^5+6 b^3 c e^4 (b d-4 a e)-384 c^4 d^3 e (b d-a e)+96 c^3 d e^2 (b d-a e)^2+16 b c^2 e^3 \left (b^2 d^2-3 a b d e+3 a^2 e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{512 c^3 e^6}\\ &=\frac{\left (128 c^4 d^4+3 b^4 e^4-32 c^3 d^2 e (5 b d-4 a e)+8 b c^2 d e^2 (2 b d-3 a e)+6 b^2 c e^3 (b d-2 a e)-2 c e \left (32 c^3 d^3-3 b^3 e^3-8 c^2 d e (2 b d-3 a e)-6 b c e^2 (b d-2 a e)\right ) x^2\right ) \sqrt{a+b x^2+c x^4}}{256 c^3 e^5}+\frac{\left (16 c^2 d^2-6 b c d e-3 b^2 e^2-6 c e (2 c d+b e) x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{96 c^2 e^3}+\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 c e}-\frac{\left (d^2 \left (c d^2-b d e+a e^2\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac{-b d+2 a e-(2 c d-b e) x^2}{\sqrt{a+b x^2+c x^4}}\right )}{e^6}-\frac{\left (256 c^5 d^5+3 b^5 e^5+6 b^3 c e^4 (b d-4 a e)-384 c^4 d^3 e (b d-a e)+96 c^3 d e^2 (b d-a e)^2+16 b c^2 e^3 \left (b^2 d^2-3 a b d e+3 a^2 e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x^2}{\sqrt{a+b x^2+c x^4}}\right )}{256 c^3 e^6}\\ &=\frac{\left (128 c^4 d^4+3 b^4 e^4-32 c^3 d^2 e (5 b d-4 a e)+8 b c^2 d e^2 (2 b d-3 a e)+6 b^2 c e^3 (b d-2 a e)-2 c e \left (32 c^3 d^3-3 b^3 e^3-8 c^2 d e (2 b d-3 a e)-6 b c e^2 (b d-2 a e)\right ) x^2\right ) \sqrt{a+b x^2+c x^4}}{256 c^3 e^5}+\frac{\left (16 c^2 d^2-6 b c d e-3 b^2 e^2-6 c e (2 c d+b e) x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{96 c^2 e^3}+\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 c e}-\frac{\left (256 c^5 d^5+3 b^5 e^5+6 b^3 c e^4 (b d-4 a e)-384 c^4 d^3 e (b d-a e)+96 c^3 d e^2 (b d-a e)^2+16 b c^2 e^3 \left (b^2 d^2-3 a b d e+3 a^2 e^2\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{512 c^{7/2} e^6}+\frac{d^2 \left (c d^2-b d e+a e^2\right )^{3/2} \tanh ^{-1}\left (\frac{b d-2 a e+(2 c d-b e) x^2}{2 \sqrt{c d^2-b d e+a e^2} \sqrt{a+b x^2+c x^4}}\right )}{2 e^6}\\ \end{align*}

Mathematica [A]  time = 1.02952, size = 545, normalized size = 1.13 \[ \frac{-\frac{240 d^2 \left ((2 c d-b e) \left (4 c e (3 a e-2 b d)-b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )+2 \sqrt{c} \left (e \sqrt{a+b x^2+c x^4} \left (-2 c e \left (4 a e-5 b d+b e x^2\right )-b^2 e^2+4 c^2 d \left (e x^2-2 d\right )\right )+8 c \left (e (a e-b d)+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{2 a e-b d+b e x^2-2 c d x^2}{2 \sqrt{a+b x^2+c x^4} \sqrt{e (a e-b d)+c d^2}}\right )\right )\right )}{c^{3/2} e^3}-\frac{90 d e \left (b^2-4 a c\right ) \left (\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )-2 \sqrt{c} \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}\right )}{c^{5/2}}+\frac{15 b e^2 \left (3 \left (b^2-4 a c\right ) \left (\frac{\left (4 a c-b^2\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{c^{3/2}}+\frac{2 \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{c}\right )-16 \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}\right )}{c^2}+1280 d^2 \left (a+b x^2+c x^4\right )^{3/2}-\frac{480 d e \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{c}+\frac{768 e^2 \left (a+b x^2+c x^4\right )^{5/2}}{c}}{7680 e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*x^2 + c*x^4)^(3/2))/(d + e*x^2),x]

[Out]

(1280*d^2*(a + b*x^2 + c*x^4)^(3/2) - (480*d*e*(b + 2*c*x^2)*(a + b*x^2 + c*x^4)^(3/2))/c + (768*e^2*(a + b*x^
2 + c*x^4)^(5/2))/c - (90*(b^2 - 4*a*c)*d*e*(-2*Sqrt[c]*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4] + (b^2 - 4*a*c)*
ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])]))/c^(5/2) + (15*b*e^2*(-16*(b + 2*c*x^2)*(a + b*x^2
+ c*x^4)^(3/2) + 3*(b^2 - 4*a*c)*((2*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/c + ((-b^2 + 4*a*c)*ArcTanh[(b +
2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/c^(3/2))))/c^2 - (240*d^2*((2*c*d - b*e)*(8*c^2*d^2 - b^2*e^2 +
4*c*e*(-2*b*d + 3*a*e))*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])] + 2*Sqrt[c]*(e*Sqrt[a + b*
x^2 + c*x^4]*(-(b^2*e^2) + 4*c^2*d*(-2*d + e*x^2) - 2*c*e*(-5*b*d + 4*a*e + b*e*x^2)) + 8*c*(c*d^2 + e*(-(b*d)
+ a*e))^(3/2)*ArcTanh[(-(b*d) + 2*a*e - 2*c*d*x^2 + b*e*x^2)/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + b*x^2
+ c*x^4])])))/(c^(3/2)*e^3))/(7680*e^3)

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Maple [B]  time = 0.049, size = 2068, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(c*x^4+b*x^2+a)^(3/2)/(e*x^2+d),x)

[Out]

3/8*d^2/e^3*a*b*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))/c^(1/2)+d^3/e^4/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2
)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x^2+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x^2+d/e)^2+(b*e-
2*c*d)/e*(x^2+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x^2+d/e))*a*b-d^4/e^5/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((
2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x^2+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x^2+d/e)^2+(b*e-2*c*d)
/e*(x^2+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x^2+d/e))*a*c+d^5/e^6/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e
^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x^2+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x^2+d/e)^2+(b*e-2*c*d)/e*(x^
2+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x^2+d/e))*b*c+7/160/e*a*b*x^2/c*(c*x^4+b*x^2+a)^(1/2)-5/32/e^2*d*a*b/c
*(c*x^4+b*x^2+a)^(1/2)+3/32/e^2*d*a*b^2/c^(3/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))-1/64/e^2*d*b^2
*x^2/c*(c*x^4+b*x^2+a)^(1/2)+1/10/e*c*x^8*(c*x^4+b*x^2+a)^(1/2)+11/80/e*b*x^6*(c*x^4+b*x^2+a)^(1/2)-5/8*d^3/e^
4*b*(c*x^4+b*x^2+a)^(1/2)+1/2*d^4/e^5*c*(c*x^4+b*x^2+a)^(1/2)-1/2*d^5/e^6*c^(3/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*
x^4+b*x^2+a)^(1/2))+2/3*d^2/e^3*a*(c*x^4+b*x^2+a)^(1/2)+1/5/e*a*x^4*(c*x^4+b*x^2+a)^(1/2)+3/256/e*b^4/c^3*(c*x
^4+b*x^2+a)^(1/2)-3/512/e*b^5/c^(7/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))+1/10/e*a^2/c*(c*x^4+b*x^
2+a)^(1/2)+1/16*d^2/e^3/c*b^2*(c*x^4+b*x^2+a)^(1/2)-1/32*d^2/e^3*b^3/c^(3/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b
*x^2+a)^(1/2))-1/4*d^3/e^4*x^2*c*(c*x^4+b*x^2+a)^(1/2)-3/4*d^3/e^4*a*c^(1/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b
*x^2+a)^(1/2))-3/16*d^3/e^4*b^2*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))/c^(1/2)+3/4*d^4/e^5*b*c^(1/2)*
ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))-1/2*d^2/e^3/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e
+c*d^2)/e^2+(b*e-2*c*d)/e*(x^2+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x^2+d/e)^2+(b*e-2*c*d)/e*(x^2+d/e)+(
a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x^2+d/e))*a^2-1/2*d^4/e^5/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c
*d^2)/e^2+(b*e-2*c*d)/e*(x^2+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x^2+d/e)^2+(b*e-2*c*d)/e*(x^2+d/e)+(a*
e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x^2+d/e))*b^2-1/2*d^6/e^7/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d
^2)/e^2+(b*e-2*c*d)/e*(x^2+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x^2+d/e)^2+(b*e-2*c*d)/e*(x^2+d/e)+(a*e^
2-b*d*e+c*d^2)/e^2)^(1/2))/(x^2+d/e))*c^2-3/32/e*a^2*b/c^(3/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))
-5/16/e^2*d*a*x^2*(c*x^4+b*x^2+a)^(1/2)+3/128/e^2*d*b^3/c^2*(c*x^4+b*x^2+a)^(1/2)-3/256/e^2*d*b^4/c^(5/2)*ln((
1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))-3/16/e^2*d*a^2*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))/c^(
1/2)-1/8/e^2*d*c*x^6*(c*x^4+b*x^2+a)^(1/2)-3/16/e^2*d*b*x^4*(c*x^4+b*x^2+a)^(1/2)+1/160/e*b^2*x^4/c*(c*x^4+b*x
^2+a)^(1/2)-1/128/e*b^3/c^2*x^2*(c*x^4+b*x^2+a)^(1/2)-5/64/e*a*b^2/c^2*(c*x^4+b*x^2+a)^(1/2)+3/64/e*a*b^3/c^(5
/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))+1/6*d^2/e^3*c*x^4*(c*x^4+b*x^2+a)^(1/2)+7/24*d^2/e^3*b*x^2
*(c*x^4+b*x^2+a)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*x^4+b*x^2+a)^(3/2)/(e*x^2+d),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*x^4+b*x^2+a)^(3/2)/(e*x^2+d),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(c*x**4+b*x**2+a)**(3/2)/(e*x**2+d),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{4} + b x^{2} + a\right )}^{\frac{3}{2}} x^{5}}{e x^{2} + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*x^4+b*x^2+a)^(3/2)/(e*x^2+d),x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2 + a)^(3/2)*x^5/(e*x^2 + d), x)